%!TEX program = xelatex
\documentclass[t,12pt,aspectratio=169]{beamer} % 16:9 宽屏比例
\usepackage{ctex} % 中文支持
\usepackage{amsmath, amssymb, bm}
\usepackage{graphicx, color}
%\usepackage{natbib}
\usepackage{hyperref}
\usepackage{booktabs}

\usepackage{tikz} % 可选：用于绘图
\usetikzlibrary{shapes,arrows,arrows.meta}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 或者更彻底地使用 Latin Modern 数学字体（需 lualatex/xelatex）
\usepackage{unicode-math}
%\setmainfont{Latin Modern Roman}
\setmathfont{Latin Modern Math}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% \usepackage{tikz} % 用于绘图
% \usetikzlibrary{shapes,arrows}
% \usetikzlibrary{circuits.ee.IEC} % 画电路
% \usetikzlibrary{arrows.meta, calc, positioning, quotes, angles, 3d}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 画函数图形
\usepackage{pgfplots} 
\pgfplotsset{compat=1.18} % 推荐设置兼容性版本

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{setspace}
\onehalfspacing
\setlength{\parskip}{1em}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\addtobeamertemplate{frametitle}{}{\vspace*{0.3em}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usetheme{Madrid}
\usecolortheme{default}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{二体问题的微分方程}
\author{五六七}
%\date{2025年10月8日}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}

% 封面页
\begin{frame}
  \titlepage
\end{frame}

% 目录页
\begin{frame}{目录}
  \tableofcontents
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{二体问题的微分方程}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[allowframebreaks]{二体问题的微分方程}

\vspace{-0.3cm}

{\color{red}例子3. 二体问题。地球绕太阳的运动轨迹。} 

\begin{center}
\includegraphics [height=0.4\textheight, width=0.6\textwidth]{pic/sun-earth-orbit.png}
\end{center}


以太阳为原点建立直角坐标系，设地球的位置为 $$\vec{r}(t) = (x(t),y(t),z(t)). $$

地球的运动速度和加速度分别为 $$\vec{r}\,'(t) = (x'(t),y'(t),z'(t)), \,\,\,\, \vec{r}\,''(t) = (x''(t),y''(t),z''(t)). $$

设地球质量为 $m_e$, 太阳质量为 $m_s$. 地球到太阳的距离为 $$|\vec{r}(t)| = \sqrt{x^2(t) + y^2(t) + z^2(t)}. $$

由万有引力定律和牛顿第二运动定律可得
$$m_e \frac{d^2 \vec{r}}{dt^2} = - G\frac{m_sm_e}{|\vec{r}(t)|^2} \frac{\vec{r}(t)}{|\vec{r}(t)|}.$$


\newpage 

写成分量的微分方程组，可得

\begin{eqnarray*}
\left\{\begin{array}{rcl}
x\,''(t) &=& - \frac{Gm_s x}{(x^2+y^2+z^2)^{3/2}}, \\ 
y\,''(t) &=& - \frac{Gm_s y}{(x^2+y^2+z^2)^{3/2}}, \\ 
z\,''(t) &=& - \frac{Gm_s z}{(x^2+y^2+z^2)^{3/2}}. 
 \end{array}\right. 
\end{eqnarray*}

\newpage 

初值条件为初始位置和初始速度，
\begin{eqnarray*}
\left\{\begin{array}{rcl}
x(t_0) &=& x_0, \\ 
y(t_0) &=& y_0, \\ 
z(t_0) &=& z_0,
\end{array}\right. 
\hspace{2cm}
\left\{\begin{array}{rcl}
x\,'(t_0) &=& u_0, \\ 
y\,'(t_0) &=& v_0, \\ 
z\,'(t_0) &=& w_0.
 \end{array}\right. 
\end{eqnarray*}



\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{求解二体问题的微分方程}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[allowframebreaks]{求解二体问题的微分方程}

\vspace{-0.3cm}

由微分方程组可得
\begin{eqnarray*}
\left\{\begin{array}{rcl}
zy'' - yz'' &=& 0, \\ 
xz'' - zx'' &=& 0, \\  
yx'' - xy'' &=& 0.  
\end{array}\right. 
\end{eqnarray*}

首次积分可得
\begin{eqnarray*}
\left\{\begin{array}{rcl}
zy' - yz' &=& C_1, \\ 
xz' - zx' &=& C_2, \\  
yx' - xy' &=& C_3. 
\end{array}\right. 
\end{eqnarray*}

分别乘以 $x,y,z$, 相加可得下式， $$ C_1x + C_2y + C_3z =0. $$


可见地球轨道在一个平面上，于是不妨设 $z=0$, 

记 $\mu = Gm_s$, 微分方程组可简化为
\begin{eqnarray*}
\left\{\begin{array}{rcl}
x\,''  + \mu x (x^2+y^2)^{-3/2} &=& 0, \\ 
y\,''  + \mu y (x^2+y^2)^{-3/2} &=& 0. 
 \end{array}\right. 
\end{eqnarray*}

两式分别乘以 $x'$ 与 $y'$ 并相加，可得 $$(x'x''+y'y'') + \mu (xx'+yy')(x^2+y^2)^{-3/2} = 0.  $$

上式是个全微分，其原函数为 $$(x')^2 + (y')^2 - 2\mu (x^2+y^2)^{-1/2} = C_4. $$


使用极坐标变量代换 $x=r\cos\theta, y=r\sin\theta$, 可得 $$(r\,')^2 + r^2(\theta\,')^2 - \frac{2\mu}{r} = C_4. $$

由 $yx\,' - xy\,'=C_3$ 可得 $r^2\theta\,' = -C_3$. 将 $\theta\,' = \frac{-C_3}{r^2}$ 代入上式，可得
$$(r\,')^2  = C_4 + \frac{2\mu}{r} - \frac{C_3^2}{r^2} 
= C_4 +\left( \frac{\mu}{C_3}\right)^2 - \left( \frac{C_3}{r} - \frac{\mu}{C_3} \right)^2. $$

由上述两式可得 $$\frac{dr}{d\theta} = \frac{r\,'}{\theta\,'} 
= \frac{\pm\sqrt{C_4 +\left( \frac{\mu}{C_3}\right)^2 - \left( \frac{C_3}{r} - \frac{\mu}{C_3} \right)^2}}{-C_3/r^2}. $$


分离变量可得 $$ \frac{d \left( \frac{C_3}{r} \right) }
{\pm \sqrt{ C_4 + \left( \frac{\mu}{C_3}\right)^2 - \left( \frac{C_3}{r} - \frac{\mu}{C_3} \right)^2 }} = d\theta. $$

根据积分公式 $\int \frac{dx}{\sqrt{a^2-x^2}} = \arccos \frac{x}{a} + C$ 可得
$$\arccos \frac{ \frac{C_3}{r} - \frac{\mu}{C_3} }{ \sqrt{ C_4 + \left( \frac{\mu}{C_3}\right)^2 } } = \theta - C_5. $$


整理可得极坐标形式的平面曲线方程 {\color{blue} $$r=\frac{p}{1+e\cos(\theta-\theta_0)}.$$} 

其中的常数为 $$e = \frac{C_3}{\mu} \sqrt{ C_4 + \left( \frac{\mu}{C_3}\right)^2 } > 0,\,\,  
p = \frac{C_3^2}{\mu}, \,\, \theta_0 -C_5. $$

\newpage 

讨论：
\begin{enumerate}
\item 当 $e<1$ 时，这是一个椭圆；
\item 当$e=1$ 时，这是一个抛物线；
\item 当 $e>1$ 时，这是一个双曲线。
\end{enumerate}


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{document}
